Module 8 Vocab Flashcards
PIE for 3 Events
Pr[A∪B∪C] = Pr[A] + Pr[B] + Pr[C] - Pr[A∩B] - Pr[B∩C] - Pr[C∩A] - Pr[A∩B∩C]
Independent
Two events A,B ⊆ Ω
A⊥B when Pr[A∩B] = Pr[A]⋅Pr[B]
Independence is ___________
A⊥B iff B⊥A
Another way to think of A∩B is…
both A and B are happening
Property Ind i
If Pr[A] = 0, then A⊥B for any B
Property Ind ii
Ω⊥E for any E
Alternate Property Ind i
∅⊥E for any E
Property Ind iii
If A⊥B, then Pr[A∪B] = 1 - (1 - Pr[A])(1 - Pr[B])
Property Ind iv
A⊥B iff ˉA⊥B iff A⊥ˉB iff ˉA⊥ˉB
Property Ind iv (words)
A and B are independent iff the complement of each event is independent of the other event and their complements are independent
Independent vs. Disjoint
disjoint events are typically not independent of each other
If A,B are independent and disjoint, then…
at least one of A,B has a probability of 0
If E⊥ˉE
then Pr[E] = 0 or 1
Any event of probability 0,1 has the property
X⊥X
If Pr[A] = 0, then A⊥B for any B
Property Ind i
A⊥B iff ˉA⊥B iff A⊥ˉB iff ˉA⊥ˉB
Property Ind iv
If A⊥B, then Pr[A∪B] = 1 - (1 - Pr[A])(1 - Pr[B])
Property Ind iii
Ω⊥E for any E
Property Ind ii
∅⊥E for any E
Alternate Property Ind i
Two Independent Bernoulli Trials
SS would be p^2, SF and FS would be pq and FF would be q^2
“at least one success and at least one failure”
count complementarily where the “bad” cases are all successes and all failures
NOT Independent 1
A⊥B, B⊥C, C⊥A but Pr[A∩B∩C] ≠ Pr[A]⋅Pr[B]⋅Pr[C]
NOT Independent 2
Pr[A∩B∩C] = Pr[A]⋅Pr[B]⋅Pr[C] but A∤B, B∤C, C∤A
(∤ means not independent)
Pairwise Independent
Events A1,…,An are PI when for any distinct i and j between 1 and n we have Ai⊥Aj
Mutually Independent
Events A1,…An are MI when for any {i1,…,ik} ⊆ [1..n]
we have Pr[Ai1,∩…∩Aik] = Pr[Ai1]⋅…⋅Pr[Aik]
Pairwise v. Mutual Independence
MI implies PI
so if something is not PI then it can’t be MI either
Generalized Ind iii
for unions of mutually independent events
Pr[A1∪…∪An] = 1 - (1 - Pr[A1]) ⋅…⋅ (1 - Pr[An])
Set intersection distributes over set union
A∩(B∪C) = (A∩B)∪(A∩C)
Complements are ________
DISJOINT
You can see ∩ as
and
like A∩B means events A and B are happening
DeMorgan’s Lemma
ˉˉˉA∪B∪C = ˉA∩ˉB∩ˉC
PIE and Unions Connection
Rolling 10 fair dice independently and rolling a fair die 10 times independently both give rise to the same probability space
When to use conditional probability
Given (Ω, Pr) we are interested in event E in a context in which we already know for sure that event U happened/is happening
E | U
E conditioned on U
Pr[E|U]
Pr[E∩U] / Pr[U] provided Pr[U] ≠ 0
Chain Rule (for 3 events)
For any events A,B,C in the same space we have:
Pr[A∩B∩C] = Pr[A] ⋅ Pr[B|A] ⋅ Pr[C|A∩B]
For any two events A,B in the same probability space, the following two statements are equivalent
(i) A⊥B
(ii) Pr[B] = 0 or (Pr[B] ≠ 0 and Pr[A|B] = Pr[A])
Chain Rule (for 2 events)
Pr[A∩B] = Pr[A] ⋅ Pr[B|A]
_________ is a particular case of the formula for ________________ probability
Independence, conditional
Generalized Chain Rule
For any events A1,…,An in the same space we have
Pr[A1∩…∩An] = Pr[A1] ⋅ Pr[A2|A1] ⋅ Pr[A3|A1∩A2] ⋅ … ⋅ Pr[An|A1∩…∩An-1]
A ⊆ B
A∩B = A
What does disjoint mean about A∩B?
A∩B = ∅
Pr[A∩B] =
Pr[A|B] ⋅ Pr[B]
In general, the branches are labeled with __________ probabilities and along each branch the _______ _____ computes the probability of the ______________.
conditional
chain rule
outcome
